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A person repairing radios finds that the time spent on the radio sets has exponential distribution with mean 20 minutes. If the radios are repaired in the order in which they come in and their arrival is approximately poisson with an average rate 15 for 8 hour day. What is the repairmen’s expected idle time each day?
(A)2.8
(B)3.2.
Correct answer is between ','. Can you explain this answer?
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Verified Answer
A person repairing radios finds that the time spent on the radio sets...
No. of hours for which the repairman remains busy in an 8-hour day
∴ Time for which repairman remains idle
= 8 − 5
= 3 hours
Most Upvoted Answer
A person repairing radios finds that the time spent on the radio sets...
Solution:
Given data:
- Time spent on radio sets has exponential distribution with mean 20 minutes.
- Radios are repaired in the order in which they come in and their arrival is approximately Poisson with an average rate of 15 for an 8-hour day.
To find: Expected idle time of repairman each day.
Let's assume X as the time taken by a repairman to repair a radio and λ as the arrival rate of radios.
From the given data,
Mean of exponential distribution = 1/λ
Mean time spent on radio sets = 20 minutes
=> 1/λ = 20
=> λ = 1/20
Also,
Rate of Poisson distribution = λ
Average rate of arrival of radios = 15 per 8-hour day
=> λ = 15/8
=> λ = 1.875
Now, we can find the probability of repairman being idle.
P(idle) = P(no radios arrive in 20 minutes)
= P(no arrivals in λX minutes)
= e^(-λX)
Expected idle time of repairman each day can be calculated as follows:
Expected idle time = ∫P(idle)dt
Where t ranges from 0 to 8 hours (or 480 minutes)
Expected idle time = ∫e^(-λX) dX
Where X ranges from 0 to infinity
Expected idle time = 1/λ = 20 minutes
Therefore, the correct option is (A) 2.8.
Given data:
- Time spent on radio sets has exponential distribution with mean 20 minutes.
- Radios are repaired in the order in which they come in and their arrival is approximately Poisson with an average rate of 15 for an 8-hour day.
To find: Expected idle time of repairman each day.
Let's assume X as the time taken by a repairman to repair a radio and λ as the arrival rate of radios.
From the given data,
Mean of exponential distribution = 1/λ
Mean time spent on radio sets = 20 minutes
=> 1/λ = 20
=> λ = 1/20
Also,
Rate of Poisson distribution = λ
Average rate of arrival of radios = 15 per 8-hour day
=> λ = 15/8
=> λ = 1.875
Now, we can find the probability of repairman being idle.
P(idle) = P(no radios arrive in 20 minutes)
= P(no arrivals in λX minutes)
= e^(-λX)
Expected idle time of repairman each day can be calculated as follows:
Expected idle time = ∫P(idle)dt
Where t ranges from 0 to 8 hours (or 480 minutes)
Expected idle time = ∫e^(-λX) dX
Where X ranges from 0 to infinity
Expected idle time = 1/λ = 20 minutes
Therefore, the correct option is (A) 2.8.
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A person repairing radios finds that the time spent on the radio sets has exponential distribution with mean 20 minutes. If the radios are repaired in the order in which they come in and their arrival is approximately poisson with an average rate 15 for 8 hour day. What is the repairmen’s expected idle time each day?(A)2.8(B)3.2.Correct answer is between ','. Can you explain this answer?
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